Let $C$ be the curve of intersection of the cylinder $\frac{x^2}{25}$ + $\frac{y^2}{9}$ = $1$ with the plane $3z = 4y$. Let $L$ be the line tangent to $C$ at the point $(0,-3,-4)$. What is the x-coordinate of the point of intersection of L and the plane $2x - 3y - 4z = 27$ ?
Don't even know where to begin this problem. Have a very shaky understanding of quadric surfaces. Can somebody walk me through this?
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Hint. Parametrise the curve - write $x$ and $y$ in terms of trig functions of $t$ to make sure the point lies on the cylinder, then find $z$ in terms of $t$ to make the point lie on the plane too. This gives a parametrisation of the curve $${\bf r}(t)=(x(t),y(t),z(t))\ .$$ A tangent to the curve is given by the derivative ${\bf r}'(t)$. This gives you an equation for $L$ in terms of $t$, substitute into the equation of the plane, find $t$ etc. Good luck!
Let $x=5\cos t,y=3\sin t$ then $z=4\sin t$ and so $$r(t)=(5\cos t,3\sin t,4\sin t)\implies t=3\pi/2\iff r=(0,-3,-4) \\ r'(t)=(-5\sin t,3\cos t,4\cos t) $$ the parallel to the line is $u=(5,0,0)$ and the equatio of $L$ is $r_1(t)=(0,-3,-4)+t(5,0,0)\implies x=5t,y=-3,z=-4 $. Finally the point of intersection is $$2(5t)-3(-3)-4(-4)=27\implies t=1/5\implies point=(1,-3,-4)$$